For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:
ΔTf<span> = (K</span>f)(<span>m)(i)
</span>where:
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3
Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol
m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg
For the problem,
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf
Tf = -1.59 celsius
Molarity = (Mass/ molar mass) x (1/ volume of solution in Litres)
Mass = Molarity x molar mass x volume of solution in Litres
Molarity of Tris = 100 mM = 0.1 M
volume of Tris sol. = 100 mL = 0.1 L
molar mass of Tris = 121.1 g/mol
Hence,
mass of Tris = Molarity of Tris x molar mass ofTris x volume of Tris solution
= 0.1 M x 121.1 g/mol x 0.1 L
= 1.211 g
mass of Tris = 1.211 g
Answer:
Hence, the wavelength of the photon associated is 1282 nm.
Explanation:
Answer:
47.8 g
Explanation:
Remember the equation for percent yield:
% yield = actual / theoretical
We're given two of the values in the question, so plug n' play:
0.945 = 45.2 / theoretical
theoretical = 47.8 g
Keep in mind you can use mass here without converting to moles because we're working with products only. If you were given a mass of reactants, you would need to convert to moles and using a balanced chemical equation find the corresponding moles of product produced.
